A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is $2$ , its base's sides have lengths of $4$ , and its base has a corner with an angle of $( pi)/6$ . What is the pyramid's surface area?

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Answer 1 · 2018-02-20T04:54:46

T S A $color(blue)(A_T = A_b + A_l = 8 + 22.6274 = 30.6274$ sq. units

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Pyramid total surface area = base area + lateral surface area

Base is a rhombus and hence $A_b = (d1 * d2)/2$ or $a^2sin theta$

Lateral Surface area $A_l = 4 * (1/2) a l$ where l is the slant height.

$A_b = 4a^2 sin (pi/6) = 8$

$l = sqrt ((a/2)^2 + h^2) = sqrt((4/2)^2 + 2^2) = 2.8284$

$A_l = 4 * (1/2) * 4 * 2.8284 = 22.6274$

$T S A$ A_T = A_b + A_l = 8 + 22.6274 = 30.6274# sq. units

A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is 2 2 , its base's sides have lengths of 4 4 , and its base has a corner with an angle of ( pi)/6 ( pi)/6 . What is the pyramid's surface area?